Question

Evaluate the definite integrals.
${\int }_2^3\frac{1}{x^2-1}dx.$

Answer 1

${\int }_2^3\frac{1}{x^2-1}dx={\left[\frac{1}{2}log\left|\frac{x-1}{x+1}\right|\right]}_2^3[\because \int \frac{dx}{x^2-a^2}=\frac{1}{2a}log\left|\frac{x-a}{x+a}\right|]=\frac{1}{2}log\left|\frac{3-1}{3+1}\right|-\frac{1}{2}log\left|\frac{2-1}{2+1}\right|=\frac{1}{2}log\left|\frac{2}{4}\right|-\frac{1}{2}log\left|\frac{1}{3}\right|=\frac{1}{2}\left[log\right(1)-log(2)-log(1)+log(3\left)\right][\because log(\frac{a}{b})=loga-logb]=\frac{1}{2}log\left(\frac{3}{2}\right)$